cosx- √2 sinx/sinx-1/2=0
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1.
\(sin^2x+cos^2x=1\Rightarrow\left(\dfrac{1}{4}\right)^2+cos^2x=1\)
\(\Rightarrow cos^2x=\dfrac{15}{16}\Rightarrow cosx=\dfrac{\sqrt{15}}{4}\)
2.
\(tanx=\dfrac{1}{3}\Rightarrow tan^2x=\dfrac{1}{9}\Rightarrow\dfrac{sin^2x}{cos^2x}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{sin^2x}{1-sin^2x}=\dfrac{1}{9}\Rightarrow9sin^2x=1-sin^2x\)
\(\Rightarrow sin^2x=\dfrac{1}{10}\Rightarrow sinx=\dfrac{\sqrt{10}}{10}\)
Giải thích các bước giải:
sin 2x=cos xsin 2x=cos x
⇔sin 2x=sin (π2−x)⇔sin 2x=sin (π2-x)
⇔⇔ ⎡⎢⎣2x=π2−x+k2π (k∈Z)2x=π−π2+x+k2π (k∈Z)[2x=π2−x+k2π (k∈Z)2x=π−π2+x+k2π (k∈Z)
⇔⇔ ⎡⎢⎣3x=π2+k2π (k∈Z)x=π2+k2π (k∈Z)[3x=π2+k2π (k∈Z)x=π2+k2π (k∈Z)
⇔⇔ ⎡⎢ ⎢⎣x=π6+k2π3 (k∈Z)x=π2+k2π (k∈Z)[x=π6+k2π3 (k∈Z)x=π2+k2π (k∈Z)
Vậy S={π6+k2π3 (k∈Z),π2+k2π (k∈Z)
\(2\left[\left(sinx+cosx+1\right)\left(sinx+cosx-1\right)\right]^2\)
\(=2\left[\left(sinx+cosx\right)^2-1\right]^2=2\left(sin^2x+cos^2x+2sinx.cosx-1\right)^2\)
\(=2\left(2sinx.cosx\right)^2=2sin^22x=1-cos4x\)
b/ \(\frac{3-4cos2a+2cos^22a-1}{3+4cos2a+2cos^22a-1}=\frac{2\left(cos^22a-2cos2a+1\right)}{2\left(cos^22a+2cos2a+1\right)}=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}\)
\(\frac{\left(1-2sin^2a-1\right)^2}{\left(2cos^2a-1+1\right)^2}=\frac{4sin^4a}{4cos^4a}=tan^4a\)
c/ \(cos^22x+sin^22x-2sin2x.cos2x+2sin3x.cosx-2sinx.cosx-sin^2x\)
\(=1-sin4x+sin4x+sin2x-sin2x-sin^2x\)
\(=1-sin^2x=cos^2x\)
\(\Leftrightarrow2-6sinx.cosx-2sinx+2cosx+2cos^2x=0\)
\(\Leftrightarrow3\left(1-2sinx.cosx\right)-2\left(sinx-cosx\right)+cos^2x-sin^2x=0\)
\(\Leftrightarrow3\left(sinx-cosx\right)^2-2\left(sinx-cosx\right)-\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx-2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\Leftrightarrow x=\frac{\pi}{4}+k\pi\\sinx-2cosx=1\left(1\right)\end{matrix}\right.\)
Xét (1) \(\Leftrightarrow\frac{1}{\sqrt{5}}sinx-\frac{2}{\sqrt{5}}cosx=\frac{1}{\sqrt{5}}\)
Đặt \(\frac{1}{\sqrt{5}}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow sinx.cosa-cosx.sina=cosa\)
\(\Leftrightarrow sin\left(x-a\right)=sin\left(\frac{\pi}{2}-a\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-a=\frac{\pi}{2}-a+k2\pi\\x-a=a+\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=2a+\frac{\pi}{2}+k2\pi\end{matrix}\right.\)